## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$x=2, y = -3 \text{ or } (2, -3)$
The system of equations is given by the following two equations: \begin{cases} \dfrac{1}{2}x+y&=-2\hspace{20pt} &(1)\3mm] x-2y&=8 \hspace{20pt} &(2) \end{cases} \text{Multiplying equation (1) by 2 so that the coefficients of y in the two equations are additive inverses gives:} x+2y=-4\hspace{20pt} (1) \text{Add the two equations to eliminate } y: \[ \begin{aligned} &x+2y=-4 \\[2mm] &x-2y=8 \ \\[3mm] \hline & 2x=4\\[2mm] \end{aligned} Divide $2$ to both sides to obtain: $x=2$ $\text{Substituting in equation (1) for }x=2$ gives: $\dfrac{1}{2} (2)+y=-2$ $y=-2-1$ $y=-3$ Therefore, the solution is: $\boxed{x=2 \hspace{20pt} y = -3}$